Answer:
1456 N
Explanation:
Given that
Frequency of the piano, f = 27.5 Hz
Entire length of the string, l = 2 m
Mass of the piano, m = 400 g
Length of the vibrating section of the string, L = 1.9 m
Tension needed, T = ?
The formula for the tension is represented as
T = 4mL²f²/ l, where
T = tension
m = mass
L = length of vibrating part
F = frequency
l = length of the whole part
If we substitute and apply the values we have Fri. The question, we would have
T = (4 * 0.4 * 1.9² * 27.5²) / 2
T = 4368.1 / 2
T = 1456 N
Thus, we could conclude that the tension needed to tune the string properly is 1456 N
Answer:
e) 120m/s
Explanation:
When the ball reaches its highest point, its velocity becomes zero, meaning
.
where
is the initial velocity.
Solving for
we get
which is the time it takes the ball to reach the highest point.
Now, after the ball has reached its highest point, it turns around and falls downwards. After time
since it had reached the highest point, the ball has traveled downwards and the velocity
it has gained is
,
and we are told that this is twice the initial velocity
; therefore,

which gives

Thus, the total time taken to reach velocity
is


This
, we are told, is 36 seconds; therefore,

and solving for
we get:



which from the options given is choice e.
The elastic potential energy of a spring is given by

where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.
The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:

In our problem, initially the spring is uncompressed, so

. Therefore, the work done by the spring when it is compressed until

is

And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
Answer:
20.0 cm
Explanation:
Here is the complete question
The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Solution
Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.
Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.
Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m
Now, P' = 1/u + 1/v
1/u = P'- 1/v
1/u = 55.0 D - 1/0.02 m
1/u = 55.0 m⁻¹ - 1/0.02 m
1/u = 55.0 m⁻¹ - 50.0 m⁻¹
1/u = 5.0 m⁻¹
u = 1/5.0 m⁻¹
u = 0.2 m
u = 20 cm
So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.